Lesson 16
Second Derivatives
September 25, 2013
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Homework 15 Questions?
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Overview
Since the last exam, we’ve covered:
1. Section 2.2 – Power rule, equations of tangent lines.
2. Section 2.3 – Product rule, quotient rule.
Today we are going to talk about how to find and interpret the derivative
of a derivative.
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Example 1 (HW #1–2)
Find the second derivative of f (x) = x 4 − 3x 3 + 5x 2 + 10.
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Example 1 (HW #1–2)
Find the second derivative of f (x) = x 4 − 3x 3 + 5x 2 + 10.
We use the power rule to find the first derivative:
f 0 (x) = 4x 3 − 9x 2 + 10x.
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Example 1 (HW #1–2)
Find the second derivative of f (x) = x 4 − 3x 3 + 5x 2 + 10.
We use the power rule to find the first derivative:
f 0 (x) = 4x 3 − 9x 2 + 10x.
Then we use the power rule again to find the second derivative:
f 00 (x) = 12x 2 − 18x + 10.
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Example 1 (HW #1–2)
Find the second derivative of f (x) = x 4 − 3x 3 + 5x 2 + 10.
We use the power rule to find the first derivative:
f 0 (x) = 4x 3 − 9x 2 + 10x.
Then we use the power rule again to find the second derivative:
f 00 (x) = 12x 2 − 18x + 10.
If we use the other notation for derivatives, the second derivative is
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d 2f
.
dx 2
Example 2 (HW #3–4)
Find the second derivative of f (x) = (x 2 + x + 1)(4 − x 2 ).
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Example 2 (HW #3–4)
Find the second derivative of f (x) = (x 2 + x + 1)(4 − x 2 ).
First, we use the product rule to find the first derivative:
f 0 (x) = (2x + 1)(4 − x 2 ) + (x 2 + x + 1)(−2x)
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Example 2 (HW #3–4)
Find the second derivative of f (x) = (x 2 + x + 1)(4 − x 2 ).
First, we use the product rule to find the first derivative:
f 0 (x) = (2x + 1)(4 − x 2 ) + (x 2 + x + 1)(−2x)
We could use the product rule again twice to find the second derivative,
but it’s easier to simplify and use the power rule.
f 0 (x) = (8x − 2x 3 + 4 − x 2 ) + (−2x 3 − 2x 2 − 2x)
= −4x 3 − 3x 2 + 6x + 4.
Now we can use the power rule to find the second derivative.
f 00 (x) = −12x 2 − 6x + 6.
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Interpretation of the second derivative
The first derivative represents the instantaneous rate of change of a
function.
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Interpretation of the second derivative
The first derivative represents the instantaneous rate of change of a
function.
The second derivative represents the rate at which the rate of change
is changing.
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Interpretation of the second derivative
The first derivative represents the instantaneous rate of change of a
function.
The second derivative represents the rate at which the rate of change
is changing.
For example, if s(t) = t 3 + 2t is the position of a car as a function of
time, then
s 0 (t) = v (t) = 3t 2 + 2
is the velocity, and
s 00 (t) = v 0 (t) = a(t) = 6t
is the acceleration.
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Interpretation of the second derivative
The first derivative represents the instantaneous rate of change of a
function.
The second derivative represents the rate at which the rate of change
is changing.
For example, if s(t) = t 3 + 2t is the position of a car as a function of
time, then
s 0 (t) = v (t) = 3t 2 + 2
is the velocity, and
s 00 (t) = v 0 (t) = a(t) = 6t
is the acceleration.
In the homework, there are some word problems that are repeats from
section 2.2. Only a few require finding the second derivative.
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Example 3 (HW #6–10)
In the previous slide, we’ve taken the population estimates for the number
of red kangaroos in Australia, and fit them to the model
P(t) = 8 +
10
t − 2000
where P(t) is the number of kangaroos, in millions, in the year t.
Q1: What is the predicted population growth rate in 2009?
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Example 3 (HW #6–10)
In the previous slide, we’ve taken the population estimates for the number
of red kangaroos in Australia, and fit them to the model
P(t) = 8 +
10
t − 2000
where P(t) is the number of kangaroos, in millions, in the year t.
Q1: What is the predicted population growth rate in 2009?
We want P 0 (2009). By the quotient rule,
P 0 (t) =
and so P 0 (2009) = −
−10
(t − 2000)2
10
= −123,457 kangaroos/year .
81
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Example 3, part 2
Our model for the population of red kangaroos in Australia is,
P(t) = 8 +
10
t − 2000
where P(t) is the number of kangaroos, in millions, in the year t.
Q2: What does the model predict was the actual population growth in
2004?
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Example 3, part 2
Our model for the population of red kangaroos in Australia is,
P(t) = 8 +
10
t − 2000
where P(t) is the number of kangaroos, in millions, in the year t.
Q2: What does the model predict was the actual population growth in
2004?
We want P(2004) − P(2003). So,
P(2004) − P(2003) = 8 +
10
10
10
−8−
=− .
4
3
12
That means the model predicts 833,333 kangaroos died in 2004.
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Example 3, part 3
Our model for the population of red kangaroos in Australia is,
P(t) = 8 +
10
t − 2000
where P(t) is the number of kangaroos, in millions, in the year t.
Q3: What does the model predict will be the population growth of
kangaroos in the long term?
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Example 3, part 3
Our model for the population of red kangaroos in Australia is,
P(t) = 8 +
10
t − 2000
where P(t) is the number of kangaroos, in millions, in the year t.
Q3: What does the model predict will be the population growth of
kangaroos in the long term?
We want lim P 0 (t). From the first part,
t→∞
lim P 0 (t) = lim −
t→∞
t→∞
10
10
= 0.
= lim − 2
2
t→∞ t − 4000t + 4000000
(t − 2000)
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Example 3, part 4
Our model for the population of red kangaroos in Australia is,
P(t) = 8 +
10
t − 2000
where P(t) is the number of kangaroos, in millions, in the year t.
Q4: What does the model predict will be the population of kangaroos in
the long run?
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Example 3, part 4
Our model for the population of red kangaroos in Australia is,
P(t) = 8 +
10
t − 2000
where P(t) is the number of kangaroos, in millions, in the year t.
Q4: What does the model predict will be the population of kangaroos in
the long run?
We want lim P(t), which is 8 million kangaroos .
t→∞
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Example 4 (HW #11–12)
A ball is thrown up in the air, and its height is given by the function
s(t) = −16t 2 + 32t, where t is in seconds and s(t) is in feet.
Q1: What is the acceleration of the ball?
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Example 4 (HW #11–12)
A ball is thrown up in the air, and its height is given by the function
s(t) = −16t 2 + 32t, where t is in seconds and s(t) is in feet.
Q1: What is the acceleration of the ball?
s 00 (t) = −32 ft/sec2 .
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Example 4 (HW #11–12)
A ball is thrown up in the air, and its height is given by the function
s(t) = −16t 2 + 32t, where t is in seconds and s(t) is in feet.
Q1: What is the acceleration of the ball?
s 00 (t) = −32 ft/sec2 .
Q2: What is the velocity when the ball hits the ground?
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Example 4 (HW #11–12)
A ball is thrown up in the air, and its height is given by the function
s(t) = −16t 2 + 32t, where t is in seconds and s(t) is in feet.
Q1: What is the acceleration of the ball?
s 00 (t) = −32 ft/sec2 .
Q2: What is the velocity when the ball hits the ground?
The ball hits the ground when s(t) = 0.
−16t 2 + 32t = 16t(−t + 2) = 0
So either t = 0 or t = 2.
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Example 4 (HW #11–12)
A ball is thrown up in the air, and its height is given by the function
s(t) = −16t 2 + 32t, where t is in seconds and s(t) is in feet.
Q1: What is the acceleration of the ball?
s 00 (t) = −32 ft/sec2 .
Q2: What is the velocity when the ball hits the ground?
The ball hits the ground when s(t) = 0.
−16t 2 + 32t = 16t(−t + 2) = 0
So either t = 0 or t = 2. The velocity when the ball hits the ground is
v (2) = −32t + 32|t=2 = −32 ft/sec .
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Review
For Friday:
1. Read p. 147–155 on the chain rule.
2. This rule can be tricky to apply if you’re not comfortable with
composing functions. Try problems #7, 8 on p.98 if you’re not sure –
we’ll show the answers on Friday.
Get ready for the quiz...!
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Quiz 6
Find the derivative. (Do not need to simplify.)
1. (x 2 + 2x + 1)(3x 2 + 1).
2.
√
1
x+√
x
2
3.
x2 + 1
.
x2 − 1
4.
3x 2 + 2x + 1
.
x 2 + 2x + 3
.
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